Answer:

Explanation:
The string vibrates in the third harmonics, n = 3
Length of the string, l = 0.36 m
Frequency of the tone produced, f = 500 Hz
The speed of sound in air is 344 m/s
Calculate the speed of sound produced by the string in the third harmonics:
The frequency of sound is given by the formula:

The magnitude of the change in momentum of the stone is about 18.4 kg.m/s

<h3>Further explanation</h3>
Let's recall Impulse formula as follows:

<em>where:</em>
<em>I = impulse on the object ( kg m/s )</em>
<em>∑F = net force acting on object ( kg m /s² = Newton )</em>
<em>t = elapsed time ( s )</em>
Let us now tackle the problem!

<u>Given:</u>
mass of ball = m = 0.500 kg
initial speed of ball = vo = 20.0 m/s
final kinetic energy = Ek = 70% Eko
<u>Asked:</u>
magnitude of the change of momentum of the stone = Δp = ?
<u>Solution:</u>
<em>Firstly, we will calculate the final speed of the ball as follows:</em>



→ <em>negative sign due to ball rebounds</em>


<em>Next, we could find the magnitude of the change of momentum of the stone as follows:</em>

![\Delta p_{stone} = - [ mv - mv_o ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%20-%20%5B%20mv%20-%20mv_o%20%5D)
![\Delta p_{stone} = m[ v_o - v ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%20m%5B%20v_o%20-%20v%20%5D)
![\Delta p_{stone} = m[ v_o + v_o\sqrt{0.7} ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%20m%5B%20v_o%20%2B%20v_o%5Csqrt%7B0.7%7D%20%5D)
![\Delta p_{stone} = mv_o [ 1 + \sqrt{0.7} ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%20mv_o%20%5B%201%20%2B%20%5Csqrt%7B0.7%7D%20%5D)
![\Delta p_{stone} = 0.500 ( 20.0 ) [ 1 + \sqrt{0.7} ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%200.500%20%28%2020.0%20%29%20%5B%201%20%2B%20%5Csqrt%7B0.7%7D%20%5D)


<h3>Learn more</h3>

<h3>Answer details</h3>
Grade: High School
Subject: Physics
Chapter: Dynamics
Answer:
a. 2 Hz b. 0.5 cycles c . 0 V
Explanation:
a. What is period of armature?
Since it takes the armature 30 seconds to complete 60 cycles, and frequency f = number of cycles/ time = 60 cycles/ 30 s = 2 cycles/ s = 2 Hz
b. How many cycles are completed in T/2 sec?
The period, T = 1/f = 1/2 Hz = 0.5 s.
So, it takes 0.5 s to complete 1 cycles. At t = T/2 = 0.5/2 = 0.25 s,
Since it takes 0.5 s to complete 1 cycle, then the number of cycles it completes in 0.25 s is 0.25/0.5 = 0.5 cycles.
c. What is the maximum emf produced when the armature completes 180° rotation?
Since the emf E = E₀sinθ and when θ = 180°, sinθ = sin180° = 0
E = E₀ × 0 = 0
E = 0
So, at 180° rotation, the maximum emf produced is 0 V.
Answer:
a) 2250 J
b) 0 J
c) 2250 J
Explanation:
a) Since, the process is isochoric
the change in internal energy

Here, n = 0.2 moles
Cv = 12.5 J/mole.K
We have to find T_f so we can use gas equation as
![\frac{P_1V_1}{P_2V_2} =\frac{T_i}{T_f}\\Since, V_1=V_2 [isochoric/process]\\\Rightarrow \frac{P_{atm}}{4P_{atm}} = \frac{300}{T_f} \\\Rightarrow T_f = 1200 K](https://tex.z-dn.net/?f=%5Cfrac%7BP_1V_1%7D%7BP_2V_2%7D%20%3D%5Cfrac%7BT_i%7D%7BT_f%7D%5C%5CSince%2C%20V_1%3DV_2%20%20%20%20%5Bisochoric%2Fprocess%5D%5C%5C%5CRightarrow%20%5Cfrac%7BP_%7Batm%7D%7D%7B4P_%7Batm%7D%7D%20%3D%20%5Cfrac%7B300%7D%7BT_f%7D%20%5C%5C%5CRightarrow%20T_f%20%3D%201200%20K)
So, 
b) Since, the process is isochoric no work shall be done.
c) By first law of thermodynamics we have

Since, Q is positive 2250 J of heat will flow into the system.
Answer:
The answer to your question is: W = 390 J
Explanation:
Work is the transfer of energy when a body is moved from one place to another.
Data
Force = 65 N
mass = 45 kg
distance = 6 meters
work = ? J
Formula
W = F x d
Process
W = 65 N x 6 m
W = 390 J